Question

A rock is thrown downward from an unknown height above the ground with an initial speed of 21m/s. It strikes the ground 8s later. Determine the initial height of the rock above the ground. The acceleration of gravity is 9m/s/s. Answer in units of m

Answer 1

Given:
initial velocity (u) = 21 m/s
time (t) = 8 seconds
acceleration (a) = 9 m/s²
distance (s) = ????????
Now,
you can use the formula,
`s=ut+ (1)/(2) a t^(2)`

Now, plug the values in the formula, and you get:


`s=21*8+ (1)/(2) *9 * 8^(2)`


`s=21*8+ (1)/(2) *9 * 64`


`s=21*8+ (1)/(2) *576`


`s=21*8+ 288`


`s=168+288`


`s= 456~meters`

SO THE INITIAL HEIGHT OF THE ROCK WAS 456 METERS.